Question

What rate of change of current in a 9.7 mH
solenoid will produce a self-induced emf of
35mV?

Answer 1

Answer ⇒ The rate of change of current is 3.608 As⁻¹

Explanation ⇒ Given conditions,

Self Induction coefficient(L) = 9.7 mH

Self Induced emf (e) = 35 mV.

Using the formula,

  e = -L × Δi/Δt

∴ Δi/Δt = -e/L

∴ Δi/Δt = -35/9.7

∴ Δi/Δt = -3.608 As⁻¹.

Hence, the rate of change of current is 3.608 As⁻¹

Hope it helps.

Answer 2

Answer:

Explanation:

Using the formula,

  e = -L × Δi/Δt

∴ Δi/Δt = -e/L

∴ Δi/Δt = -35/9.7

∴ Δi/Δt = -3.608 As⁻¹.